The speed of a projectile when it reaches its maximum height

  • Thread starter Ab17
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  • #1
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Homework Statement


The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maxi- mum height. What is the initial projection angle of the projectile?

Homework Equations


Vf2 = vi2 + 2ax


The Attempt at a Solution


I know that the speed at the top only consist of an x component and that the speed at half the height would be sqrt(vx2+vy2) . But im trying everything from two hours and not getting a solution. The answer is suppost to be 67.8 but I dont know how to get it someone help
 

Answers and Replies

  • #2
201
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The max height of the projectile is given by [itex] \frac {u^2sin^2(\theta)}{2g} [/itex] where [itex] \theta [/itex] is the angle of projection. Half the height would be [itex] \frac {u^2sin^2(\theta)}{4g} [/itex] . Find the y-component of velocity at half the height . Find the speed by adding the x and y components. Equate it to twice the x-component. Solve for [itex] \theta [/itex] . You should get your answer.
 
  • #6
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Are those the correct components?
 
  • #7
201
51
The x-component is correct ([itex] v_icos (\theta) [/itex]) the y-component is wrong. You should learn [itex] \LaTeX [/itex] else your equations are hard to read.
 
  • #9
201
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You are just giving me equations, I don't think you are putting in much effort. For the y-component [itex] v_i = v_isin(\theta) [/itex] , acceleration [itex] a = -g [/itex] and height [itex] h = \frac{v_i^2sin^2(\theta)}{4g} [/itex] What equation of motion do you use now to get [itex] v_f [/itex] ?
 
  • #10
99
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Vf2 = vi2 + 2ax
 
  • #13
201
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You have clearly made a calculation mistake , using the equation you'll get [itex] v_f^2 = \frac{v_i^2sin^2(\theta)}{2} [/itex] . The resulting equation will be [tex] 4v_i^2cos^2(\theta) = v_i^2cos^2(\theta) + \frac{v_i^2sin^2(\theta)}{2} [/tex] Solve for [itex] \theta [/itex] you'll get your answer.
 

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